A car is traveling on a level road with speed v0 at the instant when the brakes

Question

A car is traveling on a level road with speed v0 at the instant when the brakes lock, so that the tires slide rather than roll. Use the work-kinetic energy theorem to calculate the minimum stopping distance of the car in terms of v0, g, and the coefficient of kinetic friction k between the tires and the road. For formatting, just write the result rather than "d=..." (no d=). Make sure that you write the greek letter 'mu' rather than 'u'. Lastly, v initial is v subscript zero not the letter 'o'. By what factor would the minimum stopping change if (a) the coefficient of friction were doubled? (b) the initial velocity were doubled? (c) both the coefficient of friction and the initial velocity were doubled?

Explanation / Answer

A car is traveling on a level road with speed v_0 at the instant when the brakes lock, so that the tires slide rather than roll. Use the work energy theorem to calculate the minimum stopping distance of the car in terms of v_0, g, and the coefficient of kinetic friction mu_k between the tires and the road.

The work-energy theorem says that,

the net work (W) is equal to the change in kinetic energy (KE). Kinetic energy is KE = (1/2)mv².

The friction force is the only force doing work to stop the car.

This friction force is

Ff=-mg

( minus as it acts opposite of motion), and since work is force over distance, then the work done by friction is

W = -mgx.

We now may write:

W = KE
-mgx = (1/2)mv² - (1/2)mv²

Solving for x:

x =- (v² - v²) / 2g

Since final velocity (v) is zero because the car comes to rest, then the equation simplifies to:

x = -(0 - v²) / 2g
     = v² / 2g

Now,

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